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2007 IMO Shortlist Problems/G3

IMO 2007 Short list/G3

Trapezoid 17.png

The diagonals of a trapezoid $ABCD$ intersect at point $P.$

Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\angle AQD = \angle CQB,$ and line $CD$ separates points $P$ and $Q.$

Prove that $\angle BQP = \angle DAQ.$

Proof

$\angle AQD = \angle CQB \implies$

$BQ$ and $AQ$ are isogonals with respect $\angle CQD.$

$P =AC \cap BD, BC || AD \implies$

$QS || AD$ is isogonal to $QP$ with respect $\angle CQD.$

From the converse of The isogonal theorem we get

$\angle BQP = \angle SQA = \angle DAQ.$ $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

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